Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $t = \dfrac{y + 6}{-7y - 28} \div \dfrac{y^2 + 16y + 60}{y^2 + 10y} $
Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{y + 6}{-7y - 28} \times \dfrac{y^2 + 10y}{y^2 + 16y + 60} $ First factor the quadratic. $t = \dfrac{y + 6}{-7y - 28} \times \dfrac{y^2 + 10y}{(y + 6)(y + 10)} $ Then factor out any other terms. $t = \dfrac{y + 6}{-7(y + 4)} \times \dfrac{y(y + 10)}{(y + 6)(y + 10)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ (y + 6) \times y(y + 10) } { -7(y + 4) \times (y + 6)(y + 10) } $ $t = \dfrac{ y(y + 6)(y + 10)}{ -7(y + 4)(y + 6)(y + 10)} $ Notice that $(y + 10)$ and $(y + 6)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ y\cancel{(y + 6)}(y + 10)}{ -7(y + 4)\cancel{(y + 6)}(y + 10)} $ We are dividing by $y + 6$ , so $y + 6 \neq 0$ Therefore, $y \neq -6$ $t = \dfrac{ y\cancel{(y + 6)}\cancel{(y + 10)}}{ -7(y + 4)\cancel{(y + 6)}\cancel{(y + 10)}} $ We are dividing by $y + 10$ , so $y + 10 \neq 0$ Therefore, $y \neq -10$ $t = \dfrac{y}{-7(y + 4)} $ $t = \dfrac{-y}{7(y + 4)} ; \space y \neq -6 ; \space y \neq -10 $